Graphics Reference
In-Depth Information
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
x
Fig. 8.9.
A cubic interpolant between points (1, 1) and (4, 3).
d
V
out
d
t
The rate of change of
V
out
relative to
t
(i.e.
)mustequal1when
t
=0,
d
V
out
d
t
so it can be used to multiply
s
1
.When
t
=1
,
must equal 0 to attenuate
any trace of
s
1
:
d
V
out
d
t
=3
at
2
+2
bt
+
c
(8.22)
d
V
out
d
t
=1when
t
=0,and
d
V
out
d
t
but
=0when
t
= 1. Therefore
c
=1,and
3
a
+2
b
+ 1 = 0
(8.23)
But using (8.21) means that
b
=
−
2and
a
= 1. Therefore, the polynomial
V
out
has the form
V
out
=
t
3
2
t
2
+
t
−
(8.24)
Using a similar argument, one can prove that the function to blend in
s
2
equals
V
in
=
t
3
− t
2
(8.25)
Graphs of (8.24) and (8.25) are shown alongside graphs of (8.16) in Figure 8.10.
The complete interpolating function looks like
⎡
⎤
n
1
n
2
s
1
s
2
⎣
⎦
n
=[2
t
3
3
t
2
+1
2
t
3
+3
t
2
t
3
2
t
2
+
t
3
t
2
]
−
−
−
−
·
(8.26)